A heat pump operates on the ideal vapor-compression refrigeration cycle and uses refrigerant-134a as the working fluid. The condenser operates at 1000 kPa and the evaporator at 200 kPa. Determine this system's COP and the rate of heat sup - plied to the evaporator when the compressor consumes 6 kW.

QL = 24.51 kW

COP = 5.085

Explanation:

h₁ = 244.5 kj / kg, h₂ = 278.07 kj / kg, h₃ = h₄ = 107.34 kj / kg

P₁ = 1000 kPa, P₂ = 200 kPa

W = 6 kW

To determine the rate heat can use the equation

QL = m * (h₁ - h₄)

QL = W * (h₁ - h₄) / (h₂ - h₁)

QL = 6 * (244. 5 - 107.34) / (278.07 - 244.5)

QL = 24.51 kW

To determine the COP is the rate hate relation with the work done

COP = QH / W

COP = 1 + QL / Q

COP = 1 + [ 24.51 kW / 6 kW ]

COP = 5.085